Question

The initial speed of an arrow shot from a bow, at an elevation of ${30}^{\circ }$, is $15{\text{ms}}^{–1}$. Find its velocity when it hits the ground back.

• A. $7.5\text{m/sec at}{30}^{\circ }$ clockwise from horizontal
• B. $15\text{m/sec at}{30}^{\circ }$ clockwise from horizontal
• C. $7.5\text{m/sec at}{60}^{\circ }$ clockwise from horizontal
• D. $15\text{m/sec at}{60}^{\circ }$ clockwise from horizontal

Answer 1

The correct option is B $15\text{m/sec at}{30}^{\circ }$ clockwise from horizontal


Given:
Initial speed $u=15\text{m/s}$

Component along the $x-$axis

$u_x=15\cos {30}^{\circ }$

$u_x=15\frac{\sqrt{3}}{2}\text{m/s}$

Component along $y-$axis

$u_y=15\sin {30}^{\circ }=\frac{15}{2}\text{m/s}$

Time of flight is given by

$T=\frac{2u\sin \theta }{g}$

$T=\frac{2\times 15\times \left(1/2\right)}{10}=\frac{3}{2}\text{sec}$

Therefore, $v_x=u_x=15\frac{\sqrt{3}}{2}\text{m/s}(\because a_x=0)$

and $v_y=u_y-gt$

$v_y=\frac{15}{2}-10\times \frac{3}{2}=-\frac{15}{2}\text{m/s}$

$\therefore v=\sqrt{v_y^2+v_x^2}$

$v=\sqrt{{\left(\frac{15}{2}\right)}^2+{\left(\frac{15\sqrt{3}}{2}\right)}^2}=15\text{m/s}$

Now calculating the angle

tan$\theta =\frac{\left(\frac{15}{2}\right)}{\left(\frac{15\sqrt{3}}{2}\right)}=\frac{1}{\sqrt{3}}$

$\theta ={30}^0$

Hence , option $\left(A\right)$ is correct.